Did you know? You can add infinitely many positive terms and still obtain a finite total. The series 12+6+3+1.5+… approaches 24, even though the additions never literally stop. But replacing the common ratio 1/2 with 2 makes the terms grow, and the supposed “sum to infinity” no longer exists as a finite number. In JC2 H2 Mathematics, using the formula before checking convergence is a classic way to turn correct-looking algebra into an incorrect answer.
This JC2 H2 Mathematics Tutor Bukit Timah article answers a precise A-Level syllabus 9758 question: when does a geometric series converge, how do finite and infinite sums differ, and how can a learner quantify the remaining terms after an approximation? It begins with the quick condition |r|<1, then works through positive and alternating ratios, boundary cases, 52 original calculated clinics and an exam-ready error check. The mathematical cases are invented exercises, not real tuition records or financial projections.
For JC2 learners in Bukit Timah, this topic connects sequences, repeated proportions, and mathematical models of growth and decay. The 2027 Singapore–Cambridge H2 Mathematics syllabus 9758 includes finite and infinite series, the condition for convergence of a geometric series and the sum to infinity. This page focuses on the exact point where students confuse finite calculations with limiting ones; broader geometric-progression teaching has its own guide.
Quick answer: a formula is valid only with its condition
For a geometric sequence with first term a and common ratio r, the nth term is uₙ=arⁿ⁻¹. If r≠1, the finite sum of the first n terms is Sₙ=a(1−rⁿ)/(1−r). If a≠0 and |r|<1, the powers rⁿ tend to zero as n grows, and the infinite series converges to S∞=a/(1−r). The convergence condition is not optional.
- For 0<r<1: terms shrink without alternating; the infinite sum converges.
- For −1<r<0: signs alternate while magnitudes shrink; the sum still converges.
- For r=0: the first term is a and every later term is zero, so the total is a.
- For r=1 with nonzero a: Sₙ=na and there is no finite sum to infinity.
- For r=−1 with nonzero a: partial sums oscillate between a and 0 and do not converge.
- For |r|>1 with nonzero a: term magnitudes increase, so the series does not converge.
- For a=0: all terms vanish and the sum is zero, regardless of r; this is a trivial special case.
Fully worked example: 12+6+3+1.5+…
The first term is a=12 and r=1/2. Because |r|=1/2<1, the infinite sum exists: S∞=12/(1−1/2)=24. The first five terms total S₅=12+6+3+1.5+0.75=23.25. The difference between the finite sum and its infinite limit is 24−23.25=0.75.
The same difference is given by the exact remainder formula Rₙ=S∞−Sₙ=arⁿ/(1−r) for a convergent GP. With n=5, this is 12(1/2)⁵/(1/2)=0.75. Notice the exponent: after five terms the next term is ar⁵, not ar⁴. The remainder includes that next term and all those following it, so its total is not generally just the next term.
This example is about more than substituting a number into a memorised fraction. It demonstrates that “sum to infinity” is the limit of the sums of the first n terms. It also gives a way to decide whether an approximation is close enough for a stated tolerance, and it teaches why a finite sum and a single nth term have different mathematical meanings.
Where the infinite formula comes from
For r≠1, the first n terms form Sₙ=a+ar+ar²+···+arⁿ⁻¹. Multiply both sides by r and subtract the two equations. Most intermediate terms cancel, leaving (1−r)Sₙ=a(1−rⁿ). Rearranging gives the familiar finite-sum formula. This proof needs only ordinary algebra and makes the numerator and denominator signs easier to remember.
To obtain an infinite limit, examine what happens to rⁿ. If |r|<1, the term approaches zero, so Sₙ approaches a/(1−r). If r=2, rⁿ grows. If r=−2, its magnitude grows with alternating sign. If r=−1, it oscillates. In all those nonzero-first-term cases, the limiting step that produced a/(1−r) is invalid. The expression may be calculable algebraically, but it does not represent a convergent infinite sum.
A negative ratio can converge: 9−4.5+2.25−1.125+…
Here a=9 and r=−1/2. The condition is |r|=1/2<1, so S∞=9/[1−(−1/2)]=9/(3/2)=6. The first partial sums are 9, 4.5, 6.75 and 5.625. They cross above and below six because the signs alternate. Their distance from six gets smaller as more terms are included.
After four terms S₄=5.625, giving signed remainder R₄=6−5.625=0.375. After three terms S₃=6.75, giving R₃=−0.75. The remainder can be positive or negative in an alternating GP. Its absolute value describes how large the approximation error is, while its sign describes which side of the limit the partial sum lies.
Three boundary ratios: 0, 1 and −1
If a=5 and r=0, the sequence is 5,0,0,… and the infinite sum is 5. If a=5 and r=1, every term is 5 and Sₙ=5n, which has no finite limit. If a=5 and r=−1, the terms alternate 5,−5,5,−5,…; partial sums alternate between 5 and 0. There is no unique limiting sum.
It is therefore incorrect to say “every negative ratio converges” or “every ratio below one converges”. The value r=−2 is less than one, but its absolute value is two and its term magnitudes grow. The precise condition is −1<r<1 for nonzero a, equivalently |r|<1. Stating the boundary as an absolute-value inequality prevents false answers.
Finite sums exist even when the infinite series does not converge
For a=3, r=2, the first five terms are 3,6,12,24,48. They sum to S₅=93. The finite formula gives the same result, 3(1−2⁵)/(1−2)=93. This does not imply that the infinite sum exists. The terms continue doubling and the finite partial sums increase without bound. The distinction is between a fixed number of additions and a limiting value as the number grows indefinitely.
At r=1 the ordinary finite quotient formula has zero denominator. Use Sₙ=na instead, because each of the n terms equals a. A student who correctly identifies that S∞ does not exist can still be required to find a finite S₇. Do not treat “divergent” as permission to abandon an otherwise valid finite-sum question.
An error target: how many terms are enough?
When a GP converges, the exact signed remainder after n terms is Rₙ=arⁿ/(1−r). Therefore the absolute error satisfies |Rₙ|=|a| |r|ⁿ/|1−r|. If a problem asks for the smallest n making the error less than a specified ε, compare that expression with ε and test the integer boundary carefully.
For a=12 and r=1/2, |Rₙ|=24(1/2)ⁿ. To obtain error less than 0.1, require 24/2ⁿ<0.1, or 2ⁿ>240. Since 2⁷=128 and 2⁸=256, the smallest suitable integer n is 8. After eight terms the error is 24/256=0.09375. The inequality is strict, so a value exactly equal to 0.1 would not qualify.
A percentage model: contributions keep 80% of their previous size
Suppose a hypothetical process contributes 5 units, then 4, then 3.2, then 2.56, with every new contribution 80% of the preceding one. Its common ratio is r=0.8, not 80. Because |0.8|<1, the infinite sum is S∞=5/(1−0.8)=25. The first four contributions total 14.76, leaving a mathematical tail of 10.24. That tail may still be large after four terms because a ratio close to one shrinks comparatively slowly.
This interpretation belongs to the hypothetical constant-ratio model. Real financial, environmental or human systems may change their behaviour or cease after finite time. A formula based on infinitely repeated identical proportions should not be presented as a guaranteed real-world forecast. In examinations, the modelling assumptions are part of the question and should be read before applying the series.
Terms, sums and index conventions
The geometric sequence is u₁=a, u₂=ar, u₃=ar² and so on. The nth term is uₙ=arⁿ⁻¹. The first n terms added together give Sₙ. A common index error is to confuse the next omitted term arⁿ with the nth included term arⁿ⁻¹. Rewriting the first three items before a calculation is an inexpensive check of the indexing convention.
Set n=1 in the finite formula: S₁=a(1−r)/(1−r)=a for r≠1. This must agree with the first term, because a one-term sum cannot be anything else. If a proposed formula returns ar when n=1, the exponent has been shifted and the student’s answer should be corrected before a longer computation.
Comparison table: when the infinite answer exists
| First term a | Common ratio r | Infinite sum? | Result |
|---|---|---|---|
| 12 | 1/2 | Yes | 24 |
| 9 | −1/2 | Yes | 6 |
| 5 | 0 | Yes | 5 |
| 5 | 1 | No | Partial sum Sₙ=5n |
| 5 | −1 | No | Partial sums alternate 5, 0 |
| 3 | 2 | No | Term magnitudes grow |
| 0 | 2 | Yes, trivial | All terms are zero |
One-minute diagnostic
Give a learner a=8 and four possible ratios: r=1/2, −1/2, 1 and −2. Ask which are convergent before allowing a calculator. The first two are, with S∞=16 and 16/3 respectively. The other two are not. Then change only the first term to a=0 with r=−2: every term becomes zero, and the infinite sum exists trivially. That one change tests whether the student understands why the usual condition is stated for a nonzero series.
Ask next for the remainder after five terms when a=12 and r=1/2. The learner should write ar⁵/(1−r), not ar⁴. Finally ask for the fifth term separately: u₅=ar⁴. The two expressions have different roles. A student who can distinguish them orally is much less likely to commit the corresponding off-by-one error in a longer question.
Fifty-two original worked GP clinics
The clinics below use invented first terms, common ratios and finite term counts. Every case calculates Sₙ, checks convergence, and where appropriate calculates the exact S∞ and remaining signed tail. Some nonconvergent cases are deliberately mixed among convergent ones. They are mathematics practice, not actual financial projections or student test results. Work in clusters and explain the convergence decision before using a formula.
1. First term 3, ratio 1/2, 3-term sum
Question. A hypothetical geometric progression has first term a=3 and ratio r=1/2. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is 1/8. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=21/4≈5.25. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/2<1, so this nonzero GP converges. Its exact infinite sum is S∞=6≈6. The signed remainder after the first 3 terms is R3=3/4≈0.75, matching ar^3/(1−r). Always check the absolute value of r; its sign alone does not settle convergence. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
2. First term 5, ratio -1/2, 4-term sum
Question. A hypothetical geometric progression has first term a=5 and ratio r=-1/2. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 1/16. The formula Sₙ=a(1−rⁿ)/(1−r) gives S4=25/8≈3.125. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/2<1, so this nonzero GP converges. Its exact infinite sum is S∞=10/3≈3.333333. The signed remainder after the first 4 terms is R4=5/24≈0.208333, matching ar^4/(1−r). A GP with negative ratio may have a positive total even though some terms are negative. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
3. First term 7, ratio 2/3, 5-term sum
Question. A hypothetical geometric progression has first term a=7 and ratio r=2/3. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is 32/243. The formula Sₙ=a(1−rⁿ)/(1−r) gives S5=1477/81≈18.234568. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=2/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=21≈21. The signed remainder after the first 5 terms is R5=224/81≈2.765432, matching ar^5/(1−r). An exact fraction is preferable to a rounded decimal when the examination question permits an exact answer. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
4. First term 9, ratio -2/3, 6-term sum
Question. A hypothetical geometric progression has first term a=9 and ratio r=-2/3. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 64/729. The formula Sₙ=a(1−rⁿ)/(1−r) gives S6=133/27≈4.925926. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=2/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=27/5≈5.4. The signed remainder after the first 6 terms is R6=64/135≈0.474074, matching ar^6/(1−r). If the first term is positive and r is between zero and one, partial sums approach S∞ from below. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
5. First term 11, ratio 1/3, 7-term sum
Question. A hypothetical geometric progression has first term a=11 and ratio r=1/3. Calculate the sum of the first 7 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 7 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^7 is 1/2187. The formula Sₙ=a(1−rⁿ)/(1−r) gives S7=12023/729≈16.492455. This finite result is valid because exactly 7 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=33/2≈16.5. The signed remainder after the first 7 terms is R7=11/1458≈0.007545, matching ar^7/(1−r). With a negative convergent ratio, partial sums can alternate on opposite sides of the limiting sum. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
6. First term 13, ratio -1/3, 8-term sum
Question. A hypothetical geometric progression has first term a=13 and ratio r=-1/3. Calculate the sum of the first 8 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 8 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^8 is 1/6561. The formula Sₙ=a(1−rⁿ)/(1−r) gives S8=21320/2187≈9.748514. This finite result is valid because exactly 8 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=39/4≈9.75. The signed remainder after the first 8 terms is R8=13/8748≈0.001486, matching ar^8/(1−r). A finite answer never establishes that a separate infinite limit exists. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
7. First term 15, ratio 3/4, 3-term sum
Question. A hypothetical geometric progression has first term a=15 and ratio r=3/4. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is 27/64. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=555/16≈34.6875. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=3/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=60≈60. The signed remainder after the first 3 terms is R3=405/16≈25.3125, matching ar^3/(1−r). At r=1, use Sₙ=na rather than the quotient formula with a zero denominator. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
8. First term 17, ratio -3/4, 4-term sum
Question. A hypothetical geometric progression has first term a=17 and ratio r=-3/4. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 81/256. The formula Sₙ=a(1−rⁿ)/(1−r) gives S4=425/64≈6.640625. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=3/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=68/7≈9.714286. The signed remainder after the first 4 terms is R4=1377/448≈3.073661, matching ar^4/(1−r). When |r|>1 the terms do not approach zero, so the infinite series cannot converge for nonzero a. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
9. First term 19, ratio 1/4, 5-term sum
Question. A hypothetical geometric progression has first term a=19 and ratio r=1/4. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is 1/1024. The formula Sₙ=a(1−rⁿ)/(1−r) gives S5=6479/256≈25.308594. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=76/3≈25.333333. The signed remainder after the first 5 terms is R5=19/768≈0.02474, matching ar^5/(1−r). The remainder begins after the first n included terms, with the next term arⁿ. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
10. First term 21, ratio -1/4, 6-term sum
Question. A hypothetical geometric progression has first term a=21 and ratio r=-1/4. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 1/4096. The formula Sₙ=a(1−rⁿ)/(1−r) gives S6=17199/1024≈16.795898. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=84/5≈16.8. The signed remainder after the first 6 terms is R6=21/5120≈0.004102, matching ar^6/(1−r). Changing the first term multiplies every term, finite sum and convergent infinite sum by the same factor. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
11. First term 23, ratio 4/5, 7-term sum
Question. A hypothetical geometric progression has first term a=23 and ratio r=4/5. Calculate the sum of the first 7 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 7 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^7 is 16384/78125. The formula Sₙ=a(1−rⁿ)/(1−r) gives S7=1420043/15625≈90.882752. This finite result is valid because exactly 7 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=4/5<1, so this nonzero GP converges. Its exact infinite sum is S∞=115≈115. The signed remainder after the first 7 terms is R7=376832/15625≈24.117248, matching ar^7/(1−r). Before using a decimal, check the sign of the exact numerator and denominator of the proposed sum. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
12. First term 25, ratio -4/5, 8-term sum
Question. A hypothetical geometric progression has first term a=25 and ratio r=-4/5. Calculate the sum of the first 8 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 8 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^8 is 65536/390625. The formula Sₙ=a(1−rⁿ)/(1−r) gives S8=36121/3125≈11.55872. This finite result is valid because exactly 8 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=4/5<1, so this nonzero GP converges. Its exact infinite sum is S∞=125/9≈13.888889. The signed remainder after the first 8 terms is R8=65536/28125≈2.330169, matching ar^8/(1−r). The absolute remainder measures error magnitude; a signed negative remainder means the finite sum overshot its limit. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
13. First term 3, ratio 0, 3-term sum
Question. A hypothetical geometric progression has first term a=3 and ratio r=0. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is 0. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=3≈3. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=0<1, so this nonzero GP converges. Its exact infinite sum is S∞=3≈3. The signed remainder after the first 3 terms is R3=0≈0, matching ar^3/(1−r). Always check the absolute value of r; its sign alone does not settle convergence. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
14. First term 5, ratio 1, 4-term sum
Question. A hypothetical geometric progression has first term a=5 and ratio r=1. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 1. Because r=1, each term is 5, so S4=5×4=20. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=1≥1 and a≠0. The terms stay nonzero and constant. Therefore no finite S∞ exists. The finite sum S4=20 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. A GP with negative ratio may have a positive total even though some terms are negative.
15. First term 7, ratio -1, 5-term sum
Question. A hypothetical geometric progression has first term a=7 and ratio r=-1. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is -1. The formula Sₙ=a(1−rⁿ)/(1−r) gives S5=7≈7. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=1≥1 and a≠0. The terms oscillate without shrinking to zero. Therefore no finite S∞ exists. The finite sum S5=7 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. An exact fraction is preferable to a rounded decimal when the examination question permits an exact answer.
16. First term 9, ratio 2, 6-term sum
Question. A hypothetical geometric progression has first term a=9 and ratio r=2. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 64. The formula Sₙ=a(1−rⁿ)/(1−r) gives S6=567≈567. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=2≥1 and a≠0. The magnitudes increase instead of approaching zero. Therefore no finite S∞ exists. The finite sum S6=567 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. If the first term is positive and r is between zero and one, partial sums approach S∞ from below.
17. First term 11, ratio -2, 7-term sum
Question. A hypothetical geometric progression has first term a=11 and ratio r=-2. Calculate the sum of the first 7 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 7 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^7 is -128. The formula Sₙ=a(1−rⁿ)/(1−r) gives S7=473≈473. This finite result is valid because exactly 7 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=2≥1 and a≠0. The magnitudes increase instead of approaching zero. Therefore no finite S∞ exists. The finite sum S7=473 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. With a negative convergent ratio, partial sums can alternate on opposite sides of the limiting sum.
18. First term 13, ratio 3/2, 8-term sum
Question. A hypothetical geometric progression has first term a=13 and ratio r=3/2. Calculate the sum of the first 8 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 8 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^8 is 6561/256. The formula Sₙ=a(1−rⁿ)/(1−r) gives S8=81965/128≈640.351563. This finite result is valid because exactly 8 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=3/2≥1 and a≠0. The magnitudes increase instead of approaching zero. Therefore no finite S∞ exists. The finite sum S8=81965/128 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. A finite answer never establishes that a separate infinite limit exists.
19. First term 15, ratio -3/2, 3-term sum
Question. A hypothetical geometric progression has first term a=15 and ratio r=-3/2. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is -27/8. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=105/4≈26.25. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=3/2≥1 and a≠0. The magnitudes increase instead of approaching zero. Therefore no finite S∞ exists. The finite sum S3=105/4 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. At r=1, use Sₙ=na rather than the quotient formula with a zero denominator.
20. First term 17, ratio 1/2, 4-term sum
Question. A hypothetical geometric progression has first term a=17 and ratio r=1/2. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 1/16. The formula Sₙ=a(1−rⁿ)/(1−r) gives S4=255/8≈31.875. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/2<1, so this nonzero GP converges. Its exact infinite sum is S∞=34≈34. The signed remainder after the first 4 terms is R4=17/8≈2.125, matching ar^4/(1−r). When |r|>1 the terms do not approach zero, so the infinite series cannot converge for nonzero a. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
21. First term 19, ratio -1/2, 5-term sum
Question. A hypothetical geometric progression has first term a=19 and ratio r=-1/2. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is -1/32. The formula Sₙ=a(1−rⁿ)/(1−r) gives S5=209/16≈13.0625. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/2<1, so this nonzero GP converges. Its exact infinite sum is S∞=38/3≈12.666667. The signed remainder after the first 5 terms is R5=-19/48≈-0.395833, matching ar^5/(1−r). The remainder begins after the first n included terms, with the next term arⁿ. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
22. First term 21, ratio 2/3, 6-term sum
Question. A hypothetical geometric progression has first term a=21 and ratio r=2/3. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 64/729. The formula Sₙ=a(1−rⁿ)/(1−r) gives S6=4655/81≈57.469136. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=2/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=63≈63. The signed remainder after the first 6 terms is R6=448/81≈5.530864, matching ar^6/(1−r). Changing the first term multiplies every term, finite sum and convergent infinite sum by the same factor. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
23. First term 23, ratio -2/3, 7-term sum
Question. A hypothetical geometric progression has first term a=23 and ratio r=-2/3. Calculate the sum of the first 7 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 7 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^7 is -128/2187. The formula Sₙ=a(1−rⁿ)/(1−r) gives S7=10649/729≈14.607682. This finite result is valid because exactly 7 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=2/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=69/5≈13.8. The signed remainder after the first 7 terms is R7=-2944/3645≈-0.807682, matching ar^7/(1−r). Before using a decimal, check the sign of the exact numerator and denominator of the proposed sum. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
24. First term 25, ratio 1/3, 8-term sum
Question. A hypothetical geometric progression has first term a=25 and ratio r=1/3. Calculate the sum of the first 8 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 8 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^8 is 1/6561. The formula Sₙ=a(1−rⁿ)/(1−r) gives S8=82000/2187≈37.494284. This finite result is valid because exactly 8 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=75/2≈37.5. The signed remainder after the first 8 terms is R8=25/4374≈0.005716, matching ar^8/(1−r). The absolute remainder measures error magnitude; a signed negative remainder means the finite sum overshot its limit. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
25. First term 3, ratio -1/3, 3-term sum
Question. A hypothetical geometric progression has first term a=3 and ratio r=-1/3. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is -1/27. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=7/3≈2.333333. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=9/4≈2.25. The signed remainder after the first 3 terms is R3=-1/12≈-0.083333, matching ar^3/(1−r). Always check the absolute value of r; its sign alone does not settle convergence. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
26. First term 5, ratio 3/4, 4-term sum
Question. A hypothetical geometric progression has first term a=5 and ratio r=3/4. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 81/256. The formula Sₙ=a(1−rⁿ)/(1−r) gives S4=875/64≈13.671875. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=3/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=20≈20. The signed remainder after the first 4 terms is R4=405/64≈6.328125, matching ar^4/(1−r). A GP with negative ratio may have a positive total even though some terms are negative. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
27. First term 7, ratio -3/4, 5-term sum
Question. A hypothetical geometric progression has first term a=7 and ratio r=-3/4. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is -243/1024. The formula Sₙ=a(1−rⁿ)/(1−r) gives S5=1267/256≈4.949219. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=3/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=4≈4. The signed remainder after the first 5 terms is R5=-243/256≈-0.949219, matching ar^5/(1−r). An exact fraction is preferable to a rounded decimal when the examination question permits an exact answer. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
28. First term 9, ratio 1/4, 6-term sum
Question. A hypothetical geometric progression has first term a=9 and ratio r=1/4. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 1/4096. The formula Sₙ=a(1−rⁿ)/(1−r) gives S6=12285/1024≈11.99707. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=12≈12. The signed remainder after the first 6 terms is R6=3/1024≈0.00293, matching ar^6/(1−r). If the first term is positive and r is between zero and one, partial sums approach S∞ from below. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
29. First term 11, ratio -1/4, 7-term sum
Question. A hypothetical geometric progression has first term a=11 and ratio r=-1/4. Calculate the sum of the first 7 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 7 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^7 is -1/16384. The formula Sₙ=a(1−rⁿ)/(1−r) gives S7=36047/4096≈8.800537. This finite result is valid because exactly 7 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=44/5≈8.8. The signed remainder after the first 7 terms is R7=-11/20480≈-0.000537, matching ar^7/(1−r). With a negative convergent ratio, partial sums can alternate on opposite sides of the limiting sum. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
30. First term 13, ratio 4/5, 8-term sum
Question. A hypothetical geometric progression has first term a=13 and ratio r=4/5. Calculate the sum of the first 8 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 8 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^8 is 65536/390625. The formula Sₙ=a(1−rⁿ)/(1−r) gives S8=4226157/78125≈54.09481. This finite result is valid because exactly 8 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=4/5<1, so this nonzero GP converges. Its exact infinite sum is S∞=65≈65. The signed remainder after the first 8 terms is R8=851968/78125≈10.90519, matching ar^8/(1−r). A finite answer never establishes that a separate infinite limit exists. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
31. First term 15, ratio -4/5, 3-term sum
Question. A hypothetical geometric progression has first term a=15 and ratio r=-4/5. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is -64/125. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=63/5≈12.6. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=4/5<1, so this nonzero GP converges. Its exact infinite sum is S∞=25/3≈8.333333. The signed remainder after the first 3 terms is R3=-64/15≈-4.266667, matching ar^3/(1−r). At r=1, use Sₙ=na rather than the quotient formula with a zero denominator. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
32. First term 17, ratio 0, 4-term sum
Question. A hypothetical geometric progression has first term a=17 and ratio r=0. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 0. The formula Sₙ=a(1−rⁿ)/(1−r) gives S4=17≈17. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=0<1, so this nonzero GP converges. Its exact infinite sum is S∞=17≈17. The signed remainder after the first 4 terms is R4=0≈0, matching ar^4/(1−r). When |r|>1 the terms do not approach zero, so the infinite series cannot converge for nonzero a. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
33. First term 19, ratio 1, 5-term sum
Question. A hypothetical geometric progression has first term a=19 and ratio r=1. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is 1. Because r=1, each term is 19, so S5=19×5=95. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=1≥1 and a≠0. The terms stay nonzero and constant. Therefore no finite S∞ exists. The finite sum S5=95 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. The remainder begins after the first n included terms, with the next term arⁿ.
34. First term 21, ratio -1, 6-term sum
Question. A hypothetical geometric progression has first term a=21 and ratio r=-1. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 1. The formula Sₙ=a(1−rⁿ)/(1−r) gives S6=0≈0. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=1≥1 and a≠0. The terms oscillate without shrinking to zero. Therefore no finite S∞ exists. The finite sum S6=0 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. Changing the first term multiplies every term, finite sum and convergent infinite sum by the same factor.
35. First term 23, ratio 2, 7-term sum
Question. A hypothetical geometric progression has first term a=23 and ratio r=2. Calculate the sum of the first 7 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 7 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^7 is 128. The formula Sₙ=a(1−rⁿ)/(1−r) gives S7=2921≈2921. This finite result is valid because exactly 7 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=2≥1 and a≠0. The magnitudes increase instead of approaching zero. Therefore no finite S∞ exists. The finite sum S7=2921 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. Before using a decimal, check the sign of the exact numerator and denominator of the proposed sum.
36. First term 25, ratio -2, 8-term sum
Question. A hypothetical geometric progression has first term a=25 and ratio r=-2. Calculate the sum of the first 8 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 8 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^8 is 256. The formula Sₙ=a(1−rⁿ)/(1−r) gives S8=-2125≈-2125. This finite result is valid because exactly 8 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=2≥1 and a≠0. The magnitudes increase instead of approaching zero. Therefore no finite S∞ exists. The finite sum S8=-2125 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. The absolute remainder measures error magnitude; a signed negative remainder means the finite sum overshot its limit.
37. First term 3, ratio 3/2, 3-term sum
Question. A hypothetical geometric progression has first term a=3 and ratio r=3/2. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is 27/8. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=57/4≈14.25. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=3/2≥1 and a≠0. The magnitudes increase instead of approaching zero. Therefore no finite S∞ exists. The finite sum S3=57/4 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. Always check the absolute value of r; its sign alone does not settle convergence.
38. First term 5, ratio -3/2, 4-term sum
Question. A hypothetical geometric progression has first term a=5 and ratio r=-3/2. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 81/16. The formula Sₙ=a(1−rⁿ)/(1−r) gives S4=-65/8≈-8.125. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=3/2≥1 and a≠0. The magnitudes increase instead of approaching zero. Therefore no finite S∞ exists. The finite sum S4=-65/8 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. A GP with negative ratio may have a positive total even though some terms are negative.
39. First term 7, ratio 1/2, 5-term sum
Question. A hypothetical geometric progression has first term a=7 and ratio r=1/2. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is 1/32. The formula Sₙ=a(1−rⁿ)/(1−r) gives S5=217/16≈13.5625. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/2<1, so this nonzero GP converges. Its exact infinite sum is S∞=14≈14. The signed remainder after the first 5 terms is R5=7/16≈0.4375, matching ar^5/(1−r). An exact fraction is preferable to a rounded decimal when the examination question permits an exact answer. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
40. First term 9, ratio -1/2, 6-term sum
Question. A hypothetical geometric progression has first term a=9 and ratio r=-1/2. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 1/64. The formula Sₙ=a(1−rⁿ)/(1−r) gives S6=189/32≈5.90625. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/2<1, so this nonzero GP converges. Its exact infinite sum is S∞=6≈6. The signed remainder after the first 6 terms is R6=3/32≈0.09375, matching ar^6/(1−r). If the first term is positive and r is between zero and one, partial sums approach S∞ from below. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
41. First term 11, ratio 2/3, 7-term sum
Question. A hypothetical geometric progression has first term a=11 and ratio r=2/3. Calculate the sum of the first 7 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 7 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^7 is 128/2187. The formula Sₙ=a(1−rⁿ)/(1−r) gives S7=22649/729≈31.068587. This finite result is valid because exactly 7 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=2/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=33≈33. The signed remainder after the first 7 terms is R7=1408/729≈1.931413, matching ar^7/(1−r). With a negative convergent ratio, partial sums can alternate on opposite sides of the limiting sum. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
42. First term 13, ratio -2/3, 8-term sum
Question. A hypothetical geometric progression has first term a=13 and ratio r=-2/3. Calculate the sum of the first 8 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 8 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^8 is 256/6561. The formula Sₙ=a(1−rⁿ)/(1−r) gives S8=16393/2187≈7.495656. This finite result is valid because exactly 8 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=2/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=39/5≈7.8. The signed remainder after the first 8 terms is R8=3328/10935≈0.304344, matching ar^8/(1−r). A finite answer never establishes that a separate infinite limit exists. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
43. First term 15, ratio 1/3, 3-term sum
Question. A hypothetical geometric progression has first term a=15 and ratio r=1/3. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is 1/27. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=65/3≈21.666667. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=45/2≈22.5. The signed remainder after the first 3 terms is R3=5/6≈0.833333, matching ar^3/(1−r). At r=1, use Sₙ=na rather than the quotient formula with a zero denominator. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
44. First term 17, ratio -1/3, 4-term sum
Question. A hypothetical geometric progression has first term a=17 and ratio r=-1/3. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 1/81. The formula Sₙ=a(1−rⁿ)/(1−r) gives S4=340/27≈12.592593. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/3<1, so this nonzero GP converges. Its exact infinite sum is S∞=51/4≈12.75. The signed remainder after the first 4 terms is R4=17/108≈0.157407, matching ar^4/(1−r). When |r|>1 the terms do not approach zero, so the infinite series cannot converge for nonzero a. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
45. First term 19, ratio 3/4, 5-term sum
Question. A hypothetical geometric progression has first term a=19 and ratio r=3/4. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is 243/1024. The formula Sₙ=a(1−rⁿ)/(1−r) gives S5=14839/256≈57.964844. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=3/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=76≈76. The signed remainder after the first 5 terms is R5=4617/256≈18.035156, matching ar^5/(1−r). The remainder begins after the first n included terms, with the next term arⁿ. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
46. First term 21, ratio -3/4, 6-term sum
Question. A hypothetical geometric progression has first term a=21 and ratio r=-3/4. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 729/4096. The formula Sₙ=a(1−rⁿ)/(1−r) gives S6=10101/1024≈9.864258. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=3/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=12≈12. The signed remainder after the first 6 terms is R6=2187/1024≈2.135742, matching ar^6/(1−r). Changing the first term multiplies every term, finite sum and convergent infinite sum by the same factor. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
47. First term 23, ratio 1/4, 7-term sum
Question. A hypothetical geometric progression has first term a=23 and ratio r=1/4. Calculate the sum of the first 7 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 7 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^7 is 1/16384. The formula Sₙ=a(1−rⁿ)/(1−r) gives S7=125603/4096≈30.664795. This finite result is valid because exactly 7 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=92/3≈30.666667. The signed remainder after the first 7 terms is R7=23/12288≈0.001872, matching ar^7/(1−r). Before using a decimal, check the sign of the exact numerator and denominator of the proposed sum. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
48. First term 25, ratio -1/4, 8-term sum
Question. A hypothetical geometric progression has first term a=25 and ratio r=-1/4. Calculate the sum of the first 8 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 8 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^8 is 1/65536. The formula Sₙ=a(1−rⁿ)/(1−r) gives S8=327675/16384≈19.999695. This finite result is valid because exactly 8 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=1/4<1, so this nonzero GP converges. Its exact infinite sum is S∞=20≈20. The signed remainder after the first 8 terms is R8=5/16384≈0.000305, matching ar^8/(1−r). The absolute remainder measures error magnitude; a signed negative remainder means the finite sum overshot its limit. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
49. First term 3, ratio 4/5, 3-term sum
Question. A hypothetical geometric progression has first term a=3 and ratio r=4/5. Calculate the sum of the first 3 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 3 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^3 is 64/125. The formula Sₙ=a(1−rⁿ)/(1−r) gives S3=183/25≈7.32. This finite result is valid because exactly 3 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=4/5<1, so this nonzero GP converges. Its exact infinite sum is S∞=15≈15. The signed remainder after the first 3 terms is R3=192/25≈7.68, matching ar^3/(1−r). Always check the absolute value of r; its sign alone does not settle convergence. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
50. First term 5, ratio -4/5, 4-term sum
Question. A hypothetical geometric progression has first term a=5 and ratio r=-4/5. Calculate the sum of the first 4 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 4 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^4 is 256/625. The formula Sₙ=a(1−rⁿ)/(1−r) gives S4=41/25≈1.64. This finite result is valid because exactly 4 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=4/5<1, so this nonzero GP converges. Its exact infinite sum is S∞=25/9≈2.777778. The signed remainder after the first 4 terms is R4=256/225≈1.137778, matching ar^4/(1−r). A GP with negative ratio may have a positive total even though some terms are negative. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
51. First term 7, ratio 0, 5-term sum
Question. A hypothetical geometric progression has first term a=7 and ratio r=0. Calculate the sum of the first 5 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 5 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^5 is 0. The formula Sₙ=a(1−rⁿ)/(1−r) gives S5=7≈7. This finite result is valid because exactly 5 terms are being added; convergence is a separate property of the limitless process.
Convergence and remainder. Here |r|=0<1, so this nonzero GP converges. Its exact infinite sum is S∞=7≈7. The signed remainder after the first 5 terms is R5=0≈0, matching ar^5/(1−r). An exact fraction is preferable to a rounded decimal when the examination question permits an exact answer. Explain why the sign of this remainder may alternate when r is negative, while its absolute value still approaches zero.
52. First term 9, ratio 1, 6-term sum
Question. A hypothetical geometric progression has first term a=9 and ratio r=1. Calculate the sum of the first 6 terms and decide whether the associated infinite series converges. If it does, give the infinite sum and the difference between it and the first 6 terms. Do not confuse the number of terms with the exponent of the final included term.
Finite calculation. The power r^6 is 1. Because r=1, each term is 9, so S6=9×6=54. This finite result is valid because exactly 6 terms are being added; convergence is a separate property of the limitless process.
Why the infinite formula is invalid. Here |r|=1≥1 and a≠0. The terms stay nonzero and constant. Therefore no finite S∞ exists. The finite sum S6=54 is still meaningful, but simply evaluating a/(1−r) is not a legitimate infinite-sum calculation. If the first term is positive and r is between zero and one, partial sums approach S∞ from below.
Eight practical error checks
- Check |r|, not just r: −2 is less than one but has magnitude greater than one, so it does not give convergence for nonzero a.
- Check the first term: the all-zero series is a special case, even when the ratio is outside the usual convergence range.
- Check whether the question asks for Sₙ or S∞: one is finite; the other is a limit that may not exist.
- Check r=1: the normal finite quotient formula divides by zero, so use Sₙ=na.
- Check r=−1: alternating partial sums do not converge for nonzero a.
- Check the remainder index: after n included terms the first omitted term is arⁿ.
- Check the sign of the remainder: a negative convergent ratio can produce over- and undershoots.
- Check exact fractions: rounding early can obscure whether a strict error tolerance has been reached.
Twelve quick cases and answers
- 1. a=12, r=1/2: S∞=24.
- 2. a=9, r=−1/2: S∞=6.
- 3. a=8, r=−1/2: S∞=16/3.
- 4. a=5, r=0: S∞=5.
- 5. a=5, r=1: no finite S∞.
- 6. a=5, r=−1: no finite S∞.
- 7. a=3, r=2: S₅=93 but no finite S∞.
- 8. a=12, r=1/2: S₅=23.25.
- 9. For the preceding case, R₅=0.75.
- 10. a=12, r=1/2: the smallest n with |Rₙ|<0.1 is 8.
- 11. a=9, r=−1/2: S₄=5.625.
- 12. a=0, r=2: the entire zero series converges to zero.
Five-day review for JC2
Day one: decide a and r from a displayed sequence, then distinguish a term from the sum of n terms. Day two: compute Sₙ for ratios smaller, equal and larger than one, checking r=1 separately. Day three: classify convergence at r=−1, r=0, r=1 and ratios with magnitude greater than one. Day four: practise negative ratios that still converge, and interpret the sign of Rₙ. Day five: find the smallest n required for a given error tolerance, then check the integer boundary rather than rounding automatically.
A three-student tutorial can assign one learner to identify the ratio and state convergence, a second to compute Sₙ and a third to audit S∞ and the remainder. Change a sign on the next problem and rotate those roles. If a pupil can explain why r=−1/2 converges but r=−2 does not, the crucial principle has transferred beyond a single memorised example.
Frequently asked questions
Can an infinite sum of positive terms be finite?
Yes. When the ratio satisfies 0<r<1, geometric terms shrink and the partial sums approach the finite total a/(1−r).
Can a negative ratio still produce a convergent series?
Yes. When −1<r<0, the terms alternate in sign but shrink in magnitude, giving a finite limit.
Why is a/(1−r) not valid for r=2?
The formula describes the limit of a finite sum only if the series converges. With r=2 and nonzero a, the terms grow and no finite sum exists.
Is r=0 allowed?
Yes. All terms after the first are zero; the infinite sum equals the first term a.
Is the nth term the sum of the first n terms?
No. uₙ=arⁿ⁻¹ is one term, whereas Sₙ is the accumulated total of the first n terms.
How do I measure remaining error?
For a convergent GP, Rₙ=S∞−Sₙ=arⁿ/(1−r). Its absolute value measures the magnitude of the difference after n terms.
Can negative-ratio partial sums overshoot their limit?
Yes. Alternating signs can make partial sums land above and below the limit while their errors shrink in magnitude.
Is convergence tested in H2 Mathematics 9758?
Yes. The 2027 syllabus includes finite and infinite series, convergence of geometric series and their sum to infinity.
Continue the mathematics series
- Normal Conditional Tail Probabilities
- Trigonometric General Solutions and Interval Checks
- Maclaurin Quotient Series and Missing Coefficients
- Infinite Geometric Series and Convergence: current guide
Related guides include Bukit Timah H2 Sequences, Series and Recurrence, Negative Geometric Ratio and Alternation, and the Bukit Timah JC2 Tuition hub. The immutable Clementi three-pupil mathematics tutoring reference illustrates the small-group teaching model at another level.
Official curriculum source
SEAB: 2027 Singapore–Cambridge H2 Mathematics 9758 syllabus, Sequences and Series. All case values and application settings here are original mathematics exercises, not official past exam items, real tuition student outcomes or guaranteed financial results. Every infinite calculation is conditioned on the correct convergence rule.